Pages

Saturday, 30 November 2013

Constructing an AND gate as a quantum operation

Here we consider operations that map the two-bit state $\ket{a,b}$ to the one-qubit state $\ket{a\wedge b}$, for all $a,b\in\{0,1\}$, where $a\wedge b$ represents the binary logical AND operation of $a$ and $b$. We will construct $k$ matrices $A_1,\dots, A_k$ (where $k\leq 4$, and each $A_j$ is a $2 \times 4$ matrix) such that $\sum_
{j=1}^{k}A^\dagger_jA_j=I$, and whose quantum operation $\Phi$ computes the above mapping. In other words, for all $a,b\in \{0,1\}$, when $\rho=\ket{a,b}\bra{a,b}$,
\[
\Phi(\rho):=\SUM{j=1}{k}A_j\rho A^\dagger_j=\ket{a\wedge b}\bra{a\wedge b}.
\]

Now consider the four computational basis states $\ket{a,b}$, for $a,b\in\{0,1\}$, with corresponding density operators given by $\rho_i:=\ket{a,b}\bra{a,b}$ where the index can be represented in terms of $a$ and $b$ as $i=2a+b+1$. More explicitly,

Hence, the quantum operation defined in this manner does indeed compute the logical AND operation
\[
\Phi(\rho_i)=\SUM{j=1}{k}A_j\rho_i A^\dagger_j=\ket{a\wedge b}\bra{a\wedge b},
\]
where, as originally defined $\rho_i:=\ket{a,b}\bra{a,b}$ with $i=2a+b+1$ for $a,b\in\{0,1\}$.

This operation maps all basis states to pure states, but does it map all pure input states to pure output states?

Recall that $\Phi(\ket{0,0}\bra{0,0})=\ket{0}\bra{0}$ and $\Phi(\ket{1,1}\bra{1,1})=\ket{1}\bra{1}$. Moreover it can be seen that $\Phi(\ket{0,0}\bra{1,1})=\Phi(\ket{1,1}\bra{0,0})=\mathbf{0}_2$ (where $\mathbf{0}_2$ is the $2\times 2$ zero matrix) since $\Phi$ only acts nontrivially on a matrix which has some nontrivial coefficient along its main diagonal.That is, for $a,b,c,d\in\{0,1\}$, $\Phi(\ket{a,b}\bra{c,d})\neq \mathbf{0}_2$ only if $a=c$ and $b=d$. Then by the linearity of the quantum operation $\Phi$,
\[ \begin{align*}
\Phi(\rho)&=\frac{1}{2}(\Phi(\ket{0,0}\bra{0,0})+\Phi(\ket{0,0}\bra{1,1})+\Phi(\ket{1,1}\bra{0,0})+\Phi(\ket{1,1}\bra{1,1}))\\
&=\frac{1}{2}(\ket{0}\bra{0}+\ket{1}\bra{1}) \\
&=\frac{1}{2}\begin{pmatrix}1 &0 \\
0 &0 \end{pmatrix} + \frac{1}{2}\begin{pmatrix}0 &0 \\
0 &1 \end{pmatrix} \\
&=\frac{1}{2}\begin{pmatrix}1 &0 \\
0 &1 \end{pmatrix}
\end{align*}\]

Thus, since $\Phi(\rho)$ in this case is a mixed state, but $\rho$ was a pure state, the quantum operation $\Phi$ does not always take pure input states to pure output states.